Conditional Probability and
Independence
Lecturer : FATEN AL-HUSSAIN
Lecture
Contents
3-1 Introduction.
3-2 Conditional Probabilities.
3-3 Bayes’s Formula.
3-4 Independent Events.
3-5 P(.|F) Is a probability.
Summary
Problems
Theoretical Exercises
Self –Test Problems and Exercises .
Conditional Probability
If P(F)>0, then
EXAMPLE 2b:
A coin is flipped twice. Assuming that all four
points in the sample space S = {(h, h), (h, t),
(t, h), (t, t)} are equally likely, what is the
conditional probability that both flips land on
heads, given that
(a) the first flip lands on heads?
(b) at least one flip lands on heads?



Let B={(h , h)} be the event that both flips land heads.
Let F ={(h , h) , (h , t)}be the event that the first flip land
heads.
Let A={(h , h ),(h , t) ,(t , h)}be the event that at least one flip
lands heads
a)
b)
EXAMPLE 2e:
Celine is undecided as to whether to take a French
course or a chemistry course. She estimates that her
probability of receiving an A grade would be 1/2 in a
French course and 2/3 in a chemistry course. If
Celine decides to base her decision on the flip of a
fair coin, what is the probability that she gets an A in
chemistry?

Let C be the event that Celine takes chemistry.

Let A be the event that she receives an A in whatever course
she takes.
EXAMPLE 2f:
Suppose that an urn contains 8 red balls and 4
white balls. We draw 2 balls from the
urn without replacement. (a) If we assume that at
each draw each ball in the urn is
equally likely to be chosen, what is the probability
that both balls drawn are red?

Let R1 be the event that the first ball drawn is red.

Let R2 be the event that the second ball drawn is red.
BAYES’S FORMULA
Let E and F be events. We may express E as
E = EF ∪ EF c
P(E) = P(EF ) + P(EF c)
= P(E|F )P(F ) + P(E|F c)P(F c)
Main theorem:
P Bk | A 
P A|Bk   P Bk 
n
 P A|B  P B 
i 1
i
i
P  A | B   P B 
P  B | A 
c
c




P B P A| B  P B P A| B
  

EXAMPLE 3a:
An insurance company believes that people can be divided
into two classes: those who are accident prone and those who
are not. The company’s statistics show that an accident-prone
person will have an accident at some time within a fixed 1year period with probability .4, whereas this probability
decreases to .2 for a person who is not accident prone.
If we assume that 30 percent of the population is accident
prone,
what is the probability that a new policyholder will have an
accident within a year of purchasing a policy?


Let A be the event a person is accident prone
Let A1 be the event a person has an accident within 1
year
AC
=0.7
0.2
A
=0.3
A1
0.4

P A1   P AA1   P A A1
C

P(E) = P(EF ) + P(EF c)
= P(E|F )P(F ) + P(E|F c)P(F c)
P  A1   P ( A1 A) P ( A)  P ( A1 A )
C
 (0.4)( 0.3)  (0.2)( 0.7)
 0.26
Suppose that a new policyholder has an accident within a
year of purchasing a policy. What is the probability that he
or she is accident prone?
P A1 | A  P A
P A | A1  
c
c




P A  P A1 | A  P A  P A1 | A
  
(0.3)(0.4) 6


0.26
13

Alternative
A)
Alternative
B)
EXAMPLE 3c:
In answering a question on a multiple-choice test, a
student either knows the answer or guesses. Let p
be the probability that the student knows the answer
and 1 − p be the probability that the student
guesses. Assume that a student who guesses at the
answer will be correct with probability 1/m, where
m is the number of multiple-choice alternatives.
What is the conditional probability that a student
knew the answer to a question given that he or she
answered it correctly?


Let C be the event that the student answers the question
correctly .
Let K be the event that he or she actually knows the answer.
KC
=1-P
k =P
1/m
C
Alternative
EXAMPLE 3d:
A laboratory blood test is 95 percent effective in
detecting a certain disease when it is, in fact, present.
However, the test also yields a “false positive” result
for 1 percent of the healthy persons tested. (That is,
if a healthy person is tested, then, with probability
.01, the test result will imply that he or she has the
disease.) If .5 percent of the population actually has
the disease, what is the probability that a person has
the disease given that the test result is positive?


Let D be the event that the tested person has the disease .
Let E be the event that test result is positive .
DC
=.995
0.01
E
0.95
D
=.005
Alternative
INDEPENDENT EVENTS
 Two events E and F are said to be
independent if :
P(EF) = P(E)P(F)
Two coins are flipped, and all 4 outcomes
are assumed to be equally likely. If E is
the event that the first coin lands on heads
and F the event that the second lands on
tails, then E and F are independent or not?
P(EF) = P(E)P(F)
=
Proposition 4-1
If E and F are independent, then so are E and F c.
Proof.
Assume that E and F are independent.
Since E = EF ∪ EF c
EF and EF c are mutually exclusive, we have
P(E) = P(EF) + P(EF c)
= P(E).P(F)+ P(EF c )
P(EF c) =P(E) – P(E)P(F)
=p(E)[1-P(F)]
=p(E)P(F c)
Definition:
Three events E, F, and G are said to be
independent if
 P(EFG) = P(E)P(F)P(G)
 P(EF) = P(E)P(F)
 P(EG) = P(E)P(G)
 P(FG) = P(F)P(G)

Note that if E, F, and G are independent, then E will
be independent of any event formed from F and G.
For instance, E is independent of F ∪ G, since
P [E(F ∪ G)] = P(EF ∪ EG)
= P(EF) + P(EG) − P(EFG)
= P(E)P(F) + P(E)P(G) − P(E)P(FG)
= P(E)[P(F) + P(G) − P(FG)]
= P(E)P(F ∪ G)